User:Cleonis/Sandbox/Wheatstone-Foucault device

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Image 1
Wheatstone's design that displays the same effect as the Foucault pendulum.

In 1851 a paper by Charles Wheatstone was read to the Royal Society, in which Wheatstone described the device that is shown in Image 1. A transcription of the paper by Wheatstone is available at wikisource: "Note relating to M. Foucault's new mechanical proof of the Rotation of the Earth"

The helical spring acts as a heavy string; when plucked, it vibrates. The circular platform can rotate around a vertical axis. This axis will be referred to as the central axis.

In this article it will be shown that a the following set of basic physics principles suffices to account for the Foucault precession.

Newton's laws of motion
The work-energy theorem
Conservation of angular momentum

1 The Wheatstone pendulum
2 Physics principles
3 Mathematical derivations

[edit] The Wheatstone pendulum

Image 2
Version of Wheatstone's device that underlines the similarity with the Foucault pendulum

To simplify the analysis, and to underline the similarity with the Foucault pendulum, a slightly different version of the Wheatstone-Foucault device will be discussed. The device depicted in image 2 will be referred to as "the Wheatstone pendulum". The small sphere corresponds to the bob of the Foucault pendulum, the inside spring corresponds to the gravitational force that is being exerted on the bob, and the force that is exerted by the outside spring corresponds to the force that the pendulum wire exerts on the bob.

[edit] Physics principles

The assumptions for the purpose of simplifying the analysis of the Wheatstone pendulum

The Earth's gravity is not considered, as if the device is floating in an orbiting space station.
All of the mass of the bob is taken as concentrated in the midpoint of the bob
The springs are taken als massless.
The range of motion of the bob with respect to the rest state is taken to be within the limits of the Small angle approximation.

Image 3
Top: rest state. Bottom: outward displacement of pendulum bob when platform is rotating.

Image 4
Top: vibration pattern while platform is not rotating. Bottom: vibration with rotating platform.

Images 3 and 4 show a detail of the Wheatstone pendulum, in different states of motion.

Image 3:
The top side of the image shows the shape of the springs when the Wheatstone pendulum is not in motion. Both springs are equally stretched, and they are aligned. The bottom side of the image illustrates the outward displacement of the pendulum bob when the platform is rotating. The inner spring is more extended than in the rest state, the outside spring is less extended than in the rest state, and thus the springs provide the required centripetal force to make the pendulum bob move along a circular trajectory.

Image 4:
The top side of the image shows the shape of the springs when the Wheatstone pendulum is vibrating, but not rotating around its central axis. Both springs are equally stretched, and at the equilibrium point they are aligned. The bottom side of the image shows that when the pendulum bob is vibrating while the platform is rotating the midpoint of the vibration is the outward displaced point of the pendulum bob.

In these images the amplitude of the vibration is slightly smaller than the outward displacement of the bob, whereas usually the amplitude of the vibration exceeds the magnitude of the outward displacement. This does not detract from the validity of this representation for the general case. One of the properties of the Foucault pendulum is that its precession does not depend on the amplitude of the swing; when the amplitude of a Foucault pendulum decays, the rate of precession remains the same. Therefore a representation with the outward displacement roughly equal to the amplitude of the swing is just as valid; the physics principles that are at play are the same in both cases.

Let the platform be rotating counterclockwise. To underline the comparison with the Foucault pendulum, let the bob's motion in a direction tangent to the rotation be called 'east-west motion'.

[edit] Motion towards and away from the central axis

Image 5
Schematic representation of the motion of the bob as seen from a co-rotating point of view.

Image 6

At every point in time, the bob has a particular angular momentum with respect to the central axis of rotation. When the inside spring pulls the pendulum bob closer to the central axis of rotation, the centripetal force is doing mechanical work. As a consequence of the centripetal force doing work, the angular velocity of the bob changes. (Compare a spinning ice-skater who pulls her arms closer to her body to make herself spin faster.) The amount of angular acceleration can be calculated with the help of the principle of conservation of angular momentum.
When the pendulum bob is moving away from the central axis the centripetal force is doing negative work, and the bob's angular velocity decreases.

[edit] Motion tangent to the rotation

Image 7
Motion of the bob of the Wheatstone pendulum as seen from straight above the platform. Angle of the springs corresponds to 30 degrees latitude.

During a swing of the pendulum bob from west-to-east, the bob is circumnavigating the central axis faster than the equilibrium velocity, hence during a west-to-east swing the bob will swing wide. (Compare a car going round on a banked circuit, with the incline of the embankment designed for a particular speed. A car that goes faster than that will tend to climb up the embankment.) During a swing of the pendulum bob from east-to-west, the bob is circumnavigating the central axis slower than the equilibrium velocity, hence during an east-to-west swing the centripetal force will pull the bob closer to the central axis.

[edit] The dependency on the sine of the latitude

Image 8 In the case of an angle of 60 degrees with the central axis of rotation: If the two extremal points of the swing are a distance of L apart, then the motion towards and away from the central axis covers a distance of 1/2 L

In the case of the Wheatstone pendulum, just as in the case of the Foucault pendulum, the amount of precession with respect to the rotating system depends on the angle between the central axis and the latitude of the pendulum.

Image 8 illustrates the physical reason for that dependency. The amount of work that the centripetal force does is proportional to the amount of motion towards or away from the central axis.

In the case of a pendulum located at 30 degrees latitude: when the swing from one extremal point to the opposite extremal point covers a distance of L, then the motion towards (or away from) the central axis is over a distance of 1/2 L. This determines to what extend the centripetal force will affect the plane of swing. The closer to the equator, the weaker the coupling between the rotation and the vibration.

[edit] Mathematical derivations

[edit] Relations applying to motion with respect to the inertial frame

First, some relations will be derived that are designed to be a faithful representation ofthe physics taking place.

In the section 'The equation of motion for the Earthbound frame' it will be discussed how the equation of motion is set up for rotating coordinates. The advantage of using a rotating coordinate system is that in the equation of motion for rotating coordinates some of the terms drop away against each other, resulting in much simplified calculations.

Overview of symbols

r	distance to the central rotation axis
Ω	Angular velocity of the rotating system
ω	Angular velocity with respect to the inertial frame of an object
v_r	velocity in radial direction
v_t	total velocity in tangential direction
a_t	acceleration in tangential direction
a_r	acceleration in radial direction
v_t,r	velocity in tangential direction, relative to the rotating system
v_t,e	tangential velocity that corresponds to co-rotating with the rotating system, this is equilibrium velocity

[edit] Motion towards and away from the central axis

In the case of rotation, with a centripetal force that is a harmonic force, we have that velocity in radial direction results in change of angular velocity (Compare a spinning ice-skater who pulls her arms closer to her body to make herself spin faster.)

One way of finding the relation between radial velocity and angular accelation is as follows. Calculate how much force must be exerted in order to prevent angular acceleration when there is radial velocity. If that force is not provided, the amount of angular acceleration will be the same as that force requirement.

The moment of inertia: $I = m r^2 \,\!$
The angular momentum: $L = I \omega = m r^2 \omega \,\!$
Force in tangential direction $F t$
torque: $\tau = F_t r \,\!$

Let $v r$ be velocity in radial direction.
There is the following relation between change in angular momentum and torque that is being exerted.

$F_t r = \frac{dL}{dt} = \frac {d(m \omega r^2)}{dt}$

When the change in angular velocity is small compared to the change in radial distance, then ω can be treated as a constant.

$\frac {d(m \omega r^2)}{dt} = 2m \omega r \frac{dr}{dt}$

This gives the following relation:

$F_t r = 2m \omega v_r r \,\!$

With a factor r dropping away on either side of the equation, this yields the following relation between radial velocity and required tangential force:

$F_t = 2m \Omega v_r \,\!$

If that required tangential force is provided, then angular velocity remains the same, even though there is change of radial distance. When that required tangential force is not provided, then the radial velocity results in acceleration in tangential direction

$a_t = 2 \Omega v_r \,\!$

[edit] Motion tangent to the rotation

What is sought for is how acceleration in radial direction depends on tangential velocity v_t.

When the bob circumnavigates the central axis at equilibrium velocity (v_t,e), then there is no radial acceleration (a_r) As the bob is outwardly displaced, there is at each distance r to the central axis precisely the amount of centripetal force that is required to sustain circumnavigating motion.

Let the pendulum bob be swinging from east to west. Then the bob's tangential velocity is slightly smaller than the equilibrium velocity. The bob is subject the centripetal force for the equilibrium velocity F =m(v_t,e)²/r, hence there is a surplus of centripetal force. The magnitude of the surplus can be calculated by subtracting the force that would be required for the slower velocity from the actually exerted centripetal force.

The difference between those two gives the amount of radial acceleration.

$a_r = \frac{v_{t,e}^2}{r} - \frac{v_t^2}{r}$

This can also be expressed as an equation that shows how the radial acceleration depends on the tangential velocity relative to the rotating system. v_t = v_t,e + v_t,r

$\begin{align} a_r & = \frac{v_{t,e}^2}{r} - \frac{(v_{t,e} + v_{t,r})^2}{r} \\ & = \frac{v_{t,e}^2 - v_{t,e}^2 - 2v_{t,e}v_{t,r} - v_{t,r}^2}{r} \\ & = \frac{- 2v_{t,e}v_{t,r} - v_{t,r}^2}{r} \\ \end{align}$

At all distances to the central axis, v_t,e/r=Ω, so that can be substituted. When the velocity relative to the rotating system is small the above expression simplifies to:

$a_r = - 2 \Omega v_{t,r} \,\!$

[edit] Overview

In the above discussions, the following two relations have been derived in a rough manner

$\begin{cases} a_t = 2 \Omega v_r \\ a_r = -2 \Omega v_{t,r} \end{cases}$

This can be reformulated in terms of position derivatives in a rotating coordinate system with x-direction corresponding to the tangential direction, and the y-direction corresponding to radial direction.

$\begin{cases} \dfrac{d^2x}{dt^2} = 2 \Omega \dfrac{dy}{dt} \\ \dfrac{d^2y}{dt^2} = - 2 \Omega \dfrac{dx}{dt} \end{cases}$

In the next section it will be shown that the equation of motion for the Earthbound frame confirms and generalizes the above relations.

[edit] The equation of motion for the Earthbound frame

The centrifugal term and the coriolis term are computational tools that enable the equation of motion to be formulated for motion with respect to a rotating coordinate system. The coriolis term is facultative; it is only necessary when the object that is being described has a velocity with respect to the rotating system. The centrifugal term is always present, for it is proportional to the distance to the center of rotation.

In the absence of any force the following set of equations give the equation of motion for motion with respect to a rotating coordinate system. In the form presented here, the x and y coordinate are position with respect to the central axis; the central axis is the zero point of the coordinate system. Ω is the angular velocity of the rotating system.

$\begin{cases} \dfrac{d^2x}{dt^2} = m \Omega^2 x + 2 \Omega \dfrac{dy}{dt} \\ \dfrac{d^2y}{dt^2} = m \Omega^2 y - 2 \Omega \dfrac{dx}{dt} \end{cases}$

For a derivation of the centrifugal term and the coriolis term, see the fictitious force article.

[edit] The two forces in the equation of motion

In the case of a Foucault setup, such as the Foucault pendulum or the Wheatstone pendulum, two forces must be included in the equation of motion: the centripetal force that sustains the rotation, and the restoring force that sustains the vibration.

The centripetal force of the rotation, for a coordinate system with the zero point at the central axis, is given by:

$\begin{cases} F_{c,x} = - m \Omega^2 x \\ F_{c,y} = - m \Omega^2 y \end{cases}$

The restoring force acts towards the equilibrium point of the pendulum. The coordinate system can be chosen in such a way that central axis and the equilibrium point are both on the y-axis. Let the y-coordinate of the equilibrium point be called y_e

$\begin{cases} F_{r,x} = - m \omega^2 x \\ F_{r,y} = - m \omega^2 (y - y_e) \end{cases}$

The full equation of motion of the pendulum bob contains four terms: the centrifugal term, the coriolis term, an expression for the centripetal force, and an expression for the restoring force. Those four terms are vectors, and the vector sum of those terms gives the acceleration of the bob with respect to the rotating system.

$\begin{cases} \dfrac{d^2x}{dt^2} = \Omega^2 x - \Omega^2 x - \omega^2 x & \!\!\! + \, 2 \Omega \dfrac{dy}{dt}\\ \dfrac{d^2y}{dt^2} = \Omega^2 y - \Omega^2 y - \omega^2 (y - y_e) & \!\!\! - \, 2 \Omega \dfrac{dx}{dt} \end{cases}$

It is immediately clear why it is so convenient to use a co-rotating coordinate system to do the calculations: the expression for the centripetal force and the expression for the centrifugal term drop away against each other.

The fact that the centrifugal term and the expression for the centripetal force drop away against each other is not a coincidence, it is a property of the physical system.
At every distance to the center of rotation the springs provide the amount of centripetal force that is required to sustain rotation with angular velocity Ω. This happens automatically; effectively the system is self-adjusting. For example, when the angular velocity of the system increases, the springs extend and/or contract a little more until the point is reached where the springs provide the required amount of centripetal force.

Letting the centrifugal term and the expression for the centripetal force drop away against each other:

$\begin{cases} \dfrac{d^2x}{dt^2} = - \omega^2 x & \!\!\! + \, 2 \Omega \dfrac{dy}{dt} \\ \dfrac{d^2y}{dt^2} = - \omega^2 (y - y_e) & \!\!\! - \, 2 \Omega \dfrac{dx}{dt} \end{cases}$

This can be simplified further by shifting the zero point of the coordinate system. The coriolis term only contains velocity with respect to the rotating system, so it is independent of where the zero point of the coordinate system is positioned. In the following equations x and y are not the distance to the central axis but the distance to the center point of the vibration.

$\begin{cases} \dfrac{d^2x}{dt^2} = - \omega^2 x + 2 \Omega \dfrac{dy}{dt} \\ \dfrac{d^2y}{dt^2} = - \omega^2 y - 2 \Omega \dfrac{dx}{dt} \end{cases}$

Image 8
In the case of an angle of 60 degrees with the central axis of rotation: If the two extremal points of the swing are a distance of L apart, then the motion towards and away from the central axis covers a distance of 1/2 L Hence the centripetal force will do half as much work as compared with the case of a polar pendulum.

The above equations apply for only a single case, the case when the springs are vertical, which corresponds to the case of a polar Foucault pendulum.

For the general case, what needs to be included is the aspect that is illustrated in Image 8. The equations must be formulated for a plane that is perpendicular to the line that connects the upper attachment point and the equilibrium position of the bob. The restoring force of the vibration is parallel to that plane, but the other three factors, the centripetal force, the centrifugal term and the coriolis term must be resolved into a component parallel to that plane.

$\begin{cases} \dfrac{d^2x}{dt^2} = - \omega^2 x + 2 \Omega \dfrac{dy}{dt} sin(\phi) \\ \dfrac{d^2y}{dt^2} = - \omega^2 y - 2 \Omega \dfrac{dx}{dt} sin(\phi) \end{cases}$

This equation of motion decribes that the motion with respect to the rotating system does not follow a straight line. The image on the right shows a solution to the above equation of motion with a ratio of ω to Ω of 11 to 1 (Usually the ratio of ω to Ω is in the order of thousands to one). The image depicts the case of releasing the bob in such a way that on release it has no velocity with respect to the rotating system.